1. Thevinin & Norton Equivalents
Topic 14
Thevinin & Norton Equivalents
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2. Source Transformations
We add a load iL RL vL + - R vs
open circuit If
We have a circuit with just a voltage source, a series resistor and a pair of output terminals
short circuit If vL iL vs
vs/R
open circuit voltage
short circuit current -R
These represent the two limiting cases for RL
The straight line actually represents the constraint
vL vs iL for various values of RL
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Thevenin & Norton

3. Source Transformations
We add the load iL R' Is RL vL + -
Again let’s consider the two limiting cases for RL
Suppose instead we have a circuit with a current source, a parallel resistor and output terminals
open circuit If vL iL
R'is is
open circuit voltage
short circuit If
short circuit current
-R'
From the point of view of the load resistor, the two circuits are identical if
vL vs iL for various values of RL
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Thevenin & Norton

4. Source Transformations
From the terminals a-b these two circuits are indistinguishable R vs
That is, a voltage source, vs, in series with a resistor, R … b
…can always be transformed into… a R
vs/R
… a current source of value vs/R, in parallel with the same resistor. b
and, of course, vice versa!
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Thevenin & Norton

5. Source Transformations
Since most voltage sources have at least a little resistance in series with them… R vs
…they can always be modeled alternately as current sources b a R
vs/R
In practice, if R is small we treat the source as a voltage source …
… and if it is large, as a current source. b
Even so, so long as R is neither 0 nor ∞,
we can always do the transformation
to the other kind if it suits us!
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Thevenin & Norton

6. Example 4.8 4Ω 6Ω 5Ω
30Ω
20Ω
10Ω 6v
40v
Find the power associated with the 6v source and say whether it is delivering or absorbing 4Ω
12Ω 6v
19.2v 4Ω 6Ω 5Ω
30Ω
20Ω
10Ω 6v 8A
So power of
6 x .825 = 4.95 w
is absorbed by 6v source
30Ω
20Ω 6v
1.6A 4Ω 4Ω 6Ω
30Ω
10Ω 6v
32v
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Thevenin & Norton

7. Assessment 4.15 1.6Ω Use source transformations to find the voltage v
20Ω
60v v + - 6Ω 8Ω
36A
120v
Transform this pair to a current source 5Ω
Voilà!
…and this pair 6A
20Ω 5Ω 6Ω
1.6Ω 8Ω
12A
36A v + -
But why did I pick these two?
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Thevenin & Norton

8. Assessment 4.15 6A 20Ω 5Ω 6Ω 1.6Ω 8Ω 12A 36A v + -
These current sources are all driving the same node
These resistors are all in parallel
They can be replaced by a single source…
Because as well as this
…of 30A in the up direction
My eye saw this 6A 5Ω 6Ω
1.6Ω 8Ω
12A v + -
20Ω
36A
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9. Assessment 4.15 30A 2.4Ω 1.6Ω 8Ω v + - And so here as well
So the whole works reduces to this
57.6v + -
which reduces …
72v
2.4Ω
1.6Ω 8Ω v + - 6A
…to this
57.6v + -
We have 6A here 6A
and here
The second part of the problem asks how much power the 120 v source is delivering
So we have 6x(8+1.6)=57.6 volts here
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10. Back-Annotating So all we need is this current io 6A 1.6Ω 57.6v + -48 + - 6Ω 6A 8Ω 5Ω
36A
20Ω io
This part is the equivalent to the 120Ω source…
This is the same so…
20Ω
120v
…the voltages and currents are the same
All the components are in parallel so the 57.6 volts appears here
So the 120v source is delivering 3.12 A and therefore watts
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11. At Speed Original 7 calculations 20Ω 5Ω 6Ω 1.6Ω 8Ω 120v 60v 36A
2 more calculations for current and power = v + -
2 simple
calculations
9 calculations (all simple)
and
again 6A
20Ω 5Ω 6Ω
1.6Ω 8Ω
12A
36A v + -
2 calcs.
back
annotate
57.6v
48v
30A
2.4Ω
1.6Ω 8Ω v + -
2 calcs.
72v
2.4Ω
1.6Ω 8Ω v + -
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1 calc.

12. Thevenin Equivalents Its Thevenin equivalent RTh a a
vTh
RTh a b
Arbitrary resistive network containing dependent and independent sources a
From the point-of-view of terminals a-b, can always be replaced by… b
If we short-circuit them
If we open-circuit the terminals
That is:
We can find the Thevenin voltage for the arbitrary circuit by open-circuiting the terminals and measuring the voltage So
and the Thevenin resistance by short-circuiting the terminals, measuring the current, isc, and dividing it into vTh
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Thevenin & Norton

13. Thevenin Equivalents 1.4A 5Ω 4Ω 20Ω 25v 3A a b voc + - 
Find the Thevenin equivalent of the circuit shown from terminals a-b
1.6A
Find the open circuit voltage
Let’s use superposition 5Ω 4Ω
20Ω a b
voc + - 3A
25v 5Ω 4Ω
20Ω a b
voc + -
By superposition, voc=32v 3A
25v
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Thevenin & Norton

14. Thevenin Equivalents 9v + -  4A 5Ω 4Ω 20Ω 25v 3A
Find the Thevenin equivalent of the circuit shown from terminals a-b a
1.8A
16v + -
isc
.8A b
Find the short circuit current
Again with the superposition! 5Ω 4Ω
20Ω a b 3A
isc
25v 5Ω 4Ω
20Ω a b
By superposition, isc=4A 3A
25v
isc
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15. Thevenin Equivalents RTh a 32 8Ω vTh b Notice that this is just isc a
vTh/R
If we do a source transformation 4 8Ω
That is we can use the short-circuit current in parallel with the Thevenin resistance
This is actually called a Norton equivalent
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16. Thevenin & Norton Equivalents
vTh
RTh a b vo + - io
Thevenin equivalent RN iN a b vo + - io
Norton equivalent
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17. Assessing Thevenin Resistance Directly5Ω 4Ω
20Ω
25v 3A a b
voc + -
We can assess Thevenin (and therefore Norton) resistance directly
Turn off all independent sources 5Ω 4Ω
20Ω a b
RTh
Rab
Figure out the equivalent resistance looking in at the terminals
This is the same value we got by dividing vTh by isc
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18. Dependent Sources Because they are not controlled externally…
We cannot independently turn off dependent sources
…but internally, by what goes on inside the circuit
So how do we work out the equivalent resistance?
We have to do a gedanken experiment
To work out the equivalent resistance of a circuit with dependent sources from a pair of terminals, we apply, in our head…
You have to do them in your head
… or on paper or a computer …
… a test voltage source at the terminals.
Often they can’t be done in the lab!
Then we measure the current that results.
gedanken experiments were invented by Albert Einstein
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19. 24v 4A 2Ω 8Ω
3ix ix a b
Assessment 4.19
Find the Thevenin equivalent with respect to terminals a-b va 24
Use node method to find open-circuit voltage
So vTh is 8 volts
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20. Assessment 4.19 3ix it Let’s find the Thevenin Resistance 2Ω avt
Turn off the independent sources ix
24v 4A 8Ω
Apply a test voltage source to a-b b
Thus the resistance looking in at a-b is
We need to find the test current that results
Now ve do our gedanken experiment
where
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21. Assessment 4.20 60Ω 20Ω 80Ω 40Ω 160iΔ 4A iΔ a b iΔ
Find Thevenin equivalent with respect to terminals a-b ia ib
Let’s use mesh method to find voc
We have a super mesh!
KVL around super mesh
KVL around right mesh
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22. Assessment 4.20 60Ω 20Ω 80Ω 40Ω 160iΔ iΔ a b it 4A vt
Let’s find the resistance directly
Turn off the independent sources
Apply a test voltage
And solve for it
Note that
By inspection
Substituting for iΔ So
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23. Assessment 4.20
So as far as any circuitry attached to the right of terminals a-b is concerned
The whole circuit can be replaced by its Thevenin equivalent
60Ω
20Ω
80Ω
40Ω
160iΔ 4A iΔ a b
3 A a b
10Ω
30 v a b
10Ω
Or, of course, its Norton equivalent
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24. Maximum Power Transfer
We’ll just call that an arbitrary network from now on
Given an arbitrary resistive network containing dependent and independent sources and a pair of output terminals
How much power can we extract from it?
We can always transform it to its Thevenin equivalent iL
vTh
RTh a
Arbitrary Network a b vL + - RL b
Now let’s add a load…
Then the power extracted from the circuit is
… and label the voltage and current
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25. Limiting Cases iL RTh a + a RL vL vTh Arbitrary Network b - b
We know that…
Short Circuit Case:
In words: if you short the output, you max the current, but…
… there’s no voltage, so no power!
Open Circuit Case:
Again, if you open the output, you max the voltage, but…
…there’s no current, so no power!
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26. Maximum Power TransferiL
vTh
RTh a
Arbitrary Network a b RL vL + - b
Since
and
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27. Maximum power occurs where load resistance = Thevenin resistanceiL
vTh
RTh
Maximum power occurs where load resistance = Thevenin resistance a RL vL + - b
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28. Problem 4.65 Determine Thevenin equivalent with respect to a-b
1310 Ω a + ib v2
50 kΩ
100 Ω
500 µA
4x10-5v2
80ib - b
Do a source transformation
50 mv
100 Ω
50 kΩ
80ib ib v2 + -
4x105v2 a b
1310 Ω
-160 v - +
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29. Notice the Norton model might be better here
Problem 4.65
Short Circuit current
1310 Ω a + ib v2
50 kΩ
isc
100 Ω
500 µA
4x10-5v2
80ib -
=-80ib b
Notice the Norton model might be better here a b
54.6 kΩ
160 v a b
54.6 kΩ
2.84 mA
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Thevenin & Norton